As shown in my previous answer, the value of the sum that you see is $$\lim_{t\uparrow1}\sum_{n=2}^\infty (-t)^n \ln n. $$

Here is a more "manual" way to show this. Writing \begin{equation} \ln n=\int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z, \end{equation} for $t\uparrow1$ we have \begin{equation} \begin{aligned} &\sum_{n=2}^\infty (-t)^n \ln n \\ &=\sum_{n=2}^\infty (-t)^n \int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z \\ &=\int_0^\infty \frac{dz}z\sum_{n=2}^\infty (-t)^n (e^{-z}-e^{-nz}) \\ &=\frac{t^2}{1+t}\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+t e^{-z}} \\ &\to\frac12\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+e^{-z}} = \frac12\,\ln\frac\pi2. \end{aligned} \end{equation}

As shown in my previous answer, the value of the sum that you see is $$\lim_{t\uparrow1}\sum_{n=2}^\infty (-t)^n \ln n. $$

Here is a "manual" way to show this. Writing \begin{equation} \ln n=\int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z, \end{equation} for $t\uparrow1$ we have \begin{equation} \begin{aligned} &\sum_{n=2}^\infty (-t)^n \ln n \\ &=\sum_{n=2}^\infty (-t)^n \int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z \\ &=\int_0^\infty \frac{dz}z\sum_{n=2}^\infty (-t)^n (e^{-z}-e^{-nz}) \\ &=\frac{t^2}{1+t}\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+t e^{-z}} \\ &\to\frac12\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+e^{-z}} \\ &=-\int_0^1 \frac{dx}{\ln x}\,\frac{1-x}{1+x} = \frac12\,\ln\frac\pi2, \end{aligned} \end{equation} by (say) Formula 4.267.1 on p. 545 in I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Seventh Edition, 2007.